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Observation 2: rate laws and the order of reaction

We would like to understand what determines the specific dependence of the reaction rate on concentration ineach reaction. In the first case considered above, the rate depends on the concentration of the reactant to the first power. We referto this as a first order reaction . In the second case above, the rate depends on the concentration of the reactant to the secondpower, so this is called a second order reaction . There are also third order reactions , and even zeroth order reactions whose rates do not depend on the amount of the reactant. We need more observations of rate laws fordifferent reactions.

The approach used in the previous section to determine a reaction's rate law is fairly clumsy and at thispoint difficult to apply. We consider here a more systematic approach. First, consider the decomposition ofN 2 O 5 (g). 2 N 2 O 5 ( g ) 4 N O 2 ( g ) + O 2 ( g ) We can create an initial concentration of N 2 O 5 in a flask and measure the rate at which the N 2 O 5 first decomposes. We can then create a different initial concentration ofN 2 O 5 and measure the new rate at which the N 2 O 5 decomposes. By comparing these rates, we can find the order of the decomposition reaction. The rate law for decomposition ofN 2 O 5 (g) is of the general form:

Rate k [ N 2 O 5 ] m
so we need to determine the exponent m. Forexample, at 25 °C we observe that the rate of decomposition is1.4 × 10 -3 M/s when the concentration of N 2 O 5 is 0.020 M. If instead we begin with [N 2 O 5 ] = 0.010 M,we observe that the rate of decomposition is 7.0 × 10 -4 M/s. We can compare the rate from the first measurement Rate 1 to the rate from the second measurement Rate 2 . From [link] , we can write that
Rate 1 Rate 2 k [ N 2 O 5 ] 1 m k [ N 2 O 5 ] 2 m 1.4 -3 M s 7.0 -4 M s k 0.020 M m k 0.010 M m
This can be simplified on both sides of the equation to give 2.0 2.0 m Clearly, then m 1 , and the decomposition is a first order reaction. We can also thenfind the first order rate constant k for this reaction by simply plugging in one of the initial rate measurementsto [link] . We find that k 0.070 s -1 .

This approach to finding reaction order is called the method of initial rates, since it relies on fixing theconcentration at specific initial values and measuring the initial rate associated with each concentration.

So far we have considered only reactions which have a single reactant. Consider a second example of the method ofinitial rates involving the reaction of hydrogen gas and iodine gas:

H 2 ( g ) + I 2 ( g ) 2 H I ( g )
In this case, we expect to find that the rate of the reaction depends on the concentrations for both reactants.As such, we need more initial rate observations to determine the rate law. In [link] , observations are reported for the initial rate for three sets ofinitial concentrations of H 2 and I 2 .

Hydrogen gas and iodine gas initial rate data at 700k
Experiment [H 2 ] (M) [I 2 ] (M) Rate (M/sec)
1 0.10 0.10 3.00 -4
2 0.20 0.10 6.00 -4
3 0.20 0.20 1.19 -3

Following the same process we used in the N 2 O 5 example, we write the general rate law for the reaction as

Rate k [ H 2 ] n [ I 2 ] m
Rate 1 Rate 2 k [ H 2 ] 1 n [ I 2 ] 1 m k [ H 2 ] 2 n [ I 2 ] 2 m 3.00 -4 M s 6.00 -4 M s k 0.10 M m 0.10 M n k 0.20 M m 0.10 M n
0.50 0.50 m 1.00 n from which it is clear that m 1 . Similarly, we can find that n 1 . The reaction is therefore first order in each reactant and issecond order overall.
Rate k [ H 2 ] [ I 2 ]
[link] to determine the rate constant, simply by plugging in concentrationsand rate into [link] . We find that k 3.00 -2 1 M s .

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Source:  OpenStax, Concept development studies in chemistry 2012. OpenStax CNX. Aug 16, 2012 Download for free at http://legacy.cnx.org/content/col11444/1.4
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